Let $g(x)=e^x\sin(x^2)$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $e^x(\sin(x^2)+\cos(x^2))$ (Choice B) B $e^x\cos(x^2)$ (Choice C) C $2xe^x\cos(x^2)$ (Choice D) D $e^x(\sin(x^2)+2x\cos(x^2))$
Solution: $g$ is a product of a function and a composite function. Let... $u(x)=e^x$ $v(x)=\sin(x)$ $w(x)=x^2$... then $g(x)=u(x)\cdot v\Bigl(w(x)\Bigr)$. To find $g'(x)$, we will need to use the product rule and the chain rule! $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left[u(x)\cdot v\Bigl(w(x)\Bigr)\right] \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \gray{\text{Product rule}} \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=e^x$ $v'(x)=\cos(x)$ $w'(x)=2x$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot{ v\Bigl(w(x)\Bigr)}+u(x)\cdot{ v'\Bigl(w(x)\Bigr)}\cdot w'(x) \\\\ &=e^x\cdot{\sin(x^2)}+e^x\cdot{\cos(x^2)}\cdot2x \\\\ &=e^x(\sin(x^2)+2x\cos(x^2)) \end{aligned}$ In conclusion, $g'(x)=e^x(\sin(x^2)+2x\cos(x^2))$.